由題意
f(f(x+1))=(x+1)²-x-1+1=x²+x+1=f(f(x))+2x
得:f(f(1))=f(f(0))
求逆: f(1)=f(0) (1)
由原式:f(f(1))=1
故1為函數的不動點:f(1)=1
由式(1)得 f(0)=1
由題意
f(f(x+1))=(x+1)²-x-1+1=x²+x+1=f(f(x))+2x
得:f(f(1))=f(f(0))
求逆: f(1)=f(0) (1)
由原式:f(f(1))=1
故1為函數的不動點:f(1)=1
由式(1)得 f(0)=1
•
請教,由原式:f(f(1))=1 如何得到 “故1為函數的不動點:f(1)=1”
-wxcfan123-
♂
(111 bytes)
()
08/28/2024 postreply
11:40:29
•
,問得好
-大醬風度-
♂
(113 bytes)
()
08/28/2024 postreply
12:23:41
WENXUECITY.COM does not represent or guarantee the truthfulness, accuracy, or reliability of any of communications posted by other users.
Copyright ©1998-2025 wenxuecity.com All rights reserved. Privacy Statement & Terms of Use & User Privacy Protection Policy